A) \[\,\frac{2\,M}{\pi }\]
B) \[\,\frac{\,M}{\pi }\]
C) \[\frac{3\sqrt{3}\,M}{\pi }\]
D) \[\frac{3\,M}{\pi }\]
Correct Answer: D
Solution :
\[r\theta =l\] \[r=\frac{l}{\theta }\] From figure \[\sin \,\frac{\theta }{2}=\frac{x}{r}\,\,\,\,\,\Rightarrow \,\,\,x=r\,\sin \,\,\frac{\theta }{2}\] Hence new magnetic moment \[M'\,\,=\,\,m\left( 2x \right)\,\,=\,\,m.2r\,sin\,\frac{\theta }{2}\] \[=\,\,m.\frac{2l}{\theta }\sin \,\frac{\theta }{2}=\frac{2ml\sin \theta /2}{\theta }\] \[=\,\,\frac{2M\,\sin \,\,(\pi /6)}{\pi /3}\,\,=\,\,\frac{3\,M}{\pi }\]You need to login to perform this action.
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